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It appears that $$\left\lfloor\sqrt{n}\right\rfloor+\left\lfloor\sqrt{4n+1}\right\rfloor = \left\lfloor\frac{3}{2} \left\lfloor \sqrt{4n+1} \right\rfloor\right\rfloor$$ for all $n \in \mathbb{N}_{0}$, although it is not obvious to me how to prove this equality. I verified that this equality holds for $n \leq 3 \times 10^6$. I discovered this equality experimentally while researching summations involving the floor function. The integer sequence $$\left( \left\lfloor\sqrt{n}\right\rfloor+\left\lfloor\sqrt{4n+1}\right\rfloor : n \in \mathbb{N}_{0} \right) = \left(1, 3, 4, 4, 6, 6, 7, 7, 7, 9, 9, 9, 10, 10, 10, 10, \ldots \right)$$ is not currently in the On-Line Encyclopedia of Integer Sequences.

It is not clear to me how to use induction to prove this equality. Also, it is not clear to me how to use 'case analysis' to prove this equality (e.g., by considering the case whereby $4n+1$ is a perfect square). Furthermore, it is not obvious to me how to use Hermite's identity to prove this equality.

I have also considered trying to prove that $$\left\lfloor\sqrt{n}\right\rfloor+\left\lfloor\sqrt{4n+1}\right\rfloor \leq \left\lfloor\frac{3}{2} \left\lfloor \sqrt{4n+1} \right\rfloor\right\rfloor$$ and $$\left\lfloor\sqrt{n}\right\rfloor+\left\lfloor\sqrt{4n+1}\right\rfloor \geq \left\lfloor\frac{3}{2} \left\lfloor \sqrt{4n+1} \right\rfloor\right\rfloor,$$ but I have thus far only managed to prove that $$\left\lfloor\sqrt{n}\right\rfloor+\left\lfloor\sqrt{4n+1}\right\rfloor \leq \left\lfloor\frac{1}{2} \left\lfloor 3 \sqrt{4n+1} \right\rfloor\right\rfloor$$ and $$\left\lfloor \frac{3}{2} \left\lfloor \sqrt{4n+1} \right\rfloor \right\rfloor \leq \left\lfloor \sqrt{4n+1} \right\rfloor + \left\lfloor \sqrt{n + \frac{1}{4}} \right\rfloor.$$

I would be grateful to receive feedback regarding techniques which may be used to prove that $$\left\lfloor\sqrt{n}\right\rfloor+\left\lfloor\sqrt{4n+1}\right\rfloor = \left\lfloor\frac{3}{2} \left\lfloor \sqrt{4n+1} \right\rfloor\right\rfloor$$ for $n \in \mathbb{N}_{0}$, and I would be grateful to see an explicit proof of this equality.

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    $\begingroup$ Hmm…the right-hand side simplifies to $floor(\sqrt{4n+1}) + floor(\frac{1}{2} floor(\sqrt{4n+1}))$. So the conjecture reduces to $floor(\sqrt{n})= floor(\frac{1}{2} floor(\sqrt{4n+1}))$. $\endgroup$ – A.Sh Mar 22 '16 at 18:58
  • $\begingroup$ @A.Sh Thank you for your comment! It is known that for a natural number $\eta$ and $\alpha, \beta \in \mathbb{R}$, $\left\lfloor \frac{\lfloor \beta / \alpha \rfloor}{\eta} \right\rfloor = \lfloor \frac{\beta}{\alpha \eta} \rfloor$. Therefore, $\left\lfloor \frac{\lfloor \sqrt{4n+1} \rfloor}{2} \right\rfloor$ is equal to $\left\lfloor \frac{\sqrt{4n+1}}{2} \right\rfloor$. So it is clear that the problem reduces to proving that $\lfloor \sqrt{n} \rfloor = \left\lfloor \sqrt{n + \frac{1}{4}} \right\rfloor$. $\endgroup$ – John M. Campbell Mar 22 '16 at 19:09
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Following on from your comment, suppose that

$$\left\lfloor\sqrt{n}\right\rfloor<\left\lfloor\sqrt{n+\frac14}\right\rfloor\;,$$

and let $m=\left\lfloor\sqrt{n+\frac14}\right\rfloor$. Then

$$\sqrt{n}<m\le\sqrt{n+\frac14}\;,$$

so

$$n<m^2\le n+\frac14\;.$$

Since $n$ and $m^2$ are integers, this is impossible.

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